The value of lim _{x rightarrow 0} frac{1}{x} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $L = \lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$.Using the property $

Vedclass · Accepted Answer

$e^{\sin ^{2} y}$

Vedclass · Answer

(A) Let $L = \lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$.Using the property $\int_{b}^{a} f(t) dt = -\int_{a}^{b} f(t) dt$, we can rewrite the expression as:$L = \lim _{x}$ ${\rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t + \int_{a}^{x+y} e^{\sin ^{2} t} d t\right]$$L = \lim _{x \rightarrow 0} \frac{1}{x} \int_{y}^{x+y} e^{\sin ^{2} t} d t$.Since this is a $\frac{0}{0}$ form, we apply $L$'$H$ôpital's Rule and the Leibniz Integral Rule:$L = \lim _{x \rightarrow 0} \frac{\frac{d}{dx} \int_{y}^{x+y} e^{\sin ^{2} t} d t}{\frac{d}{dx} (x)}$$L = \lim _{x}$ ${\rightarrow 0} \frac{e^{\sin ^{2}(x+y)} \cdot \frac{d}{dx}(x+y) - e^{\sin ^{2}(y)} \cdot \frac{d}{dx}(y)}{1}$$L = \lim _{x \rightarrow 0} e^{\sin ^{2}(x+y)} \cdot (1) - 0 = e^{\sin ^{2} y}$.

The value of $\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$ is equal to

Similar Questions

If $f(x) = \int_{x^2}^{x^4} \sin \sqrt{t} \, dt$,then $f'(x)$ equals

The value of $\int_{0}^{1} x^{2}(1-x^{2})^{3/2} dx$ is

The angle of intersection between the curves $y = \int\limits_{x^2}^{x^3} \sqrt{5 - t^2} \, dt$ and the $x$-axis is (where $x \neq 0$):

$\int_0^\pi (\sin^3 x + \cos^2 x)^2 dx = $

$\int_0^\pi x \sin^7 x \cos^6 x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module